Fubini's theorem states that if you have a double integral over a function $f(x,y)$, then you can compute the integral as an iterated integral, if $f(x,y)$ is in $\mathcal{L}^1$. But to find out if $f$ is in $\mathcal{L}^1$ you need to compute the double integral.

What am I missing? The examples I found all apply Fubini's theorem without checking that $f(x,y)$ is in $\mathcal{L}^1$. Many thanks for any clarification!

You're right. It is a very rare situation that you know in advance that a function is integrable with respect to the product measure - it is often hard enough to show that a given function is measurable on the product spaces. My take on this is that Fubini's theorem is simply the best possible theoretical result on product measures and in the vast majority of the situations you only have to deal with $\sigma$-finite measure spaces anyway, so that Tonelli applies and the trick suggested by Nuno (or some variation of it) works.
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t.b.Jan 13 '11 at 17:54

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You don't have to compute the double integral; you just have to bound it. For example if both of the underlying measure spaces are finite and f is bounded then this is automatically true. Similarly if you are computing an integral over R^2 it suffices to bound f on a sequence of compact subsets of R^2, say concentric circles or unit squares.
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Qiaochu YuanJan 13 '11 at 17:55

Thank you, Qiaochu Yuan. I would like to accept your comment as the answer to my question, if you don't mind re-posting it as answer.
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Rudy the ReindeerJan 13 '11 at 20:04

2 Answers
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The comment made by Qiaochu Yuan answered my question but because it's a comment I can't accept it and close this question.

So hereby I declare this question as solved. Thanks to all for your help.

Here is the comment:

You don't have to compute the double
integral; you just have to bound it.
For example if both of the underlying
measure spaces are finite and f is
bounded then this is automatically
true. Similarly if you are computing
an integral over R^2 it suffices to
bound f on a sequence of compact
subsets of R^2, say concentric circles
or unit squares.

No, not really. This is a consequence of $f \in L^{1}(X \times Y)$ (at least if $X$ and $Y$ are complete). If $X$ and $Y$ fail to be complete then it is a bit more subtle, but in essence still true. See e.g. the books of Halmos or Royden
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t.b.Jan 13 '11 at 17:47